If $n$ is composite, then $\phi(n) < n-1$: hence, there is at least one number $d$ which does not divide $\phi(n)$ but divides$(n-1)$. We shall call $d$ the totient divisor of $n$. *The purist will say that totient non-divisor is a more appropriate name but for the sake of simplicity we shall stay with totient divisor*.

Let $\tau(n)$ be the number of totient divisors of $n$, i.e.,

\begin{equation} \tau(n) = \# \{d : d\mid (n-1), \ d \nmid \phi(n) \}. \end{equation}

Trivially, if $p$ is a prime then, $\tau(p) = 0$ and $\tau(p+1) = 1$. I observed the following congruences and I am looking for a proof or disproof of them:

\begin{equation} \tau(4n+3) \equiv 0 (\textrm{mod}\ 2) \end{equation}

\begin{equation} \tau(8n+5) \equiv 0 (\textrm{mod}\ 3) \end{equation}

**Motivation**:
Notice the resemblance between the above congruences and Ramanujan's partition congruences

\begin{equation} p(5n+4) \equiv 0 (\textrm{mod}\ 5) \end{equation}

\begin{equation} p(7n+5) \equiv 0 (\textrm{mod}\ 7) \end{equation}

\begin{equation} p(11n+6) \equiv 0 (\textrm{mod}\ 11) \end{equation}

**Note**: This may be related to another question on totient divisors.